Is $f^*E$ (the pullback of a smooth bundle) an embedded submanifold of $N\times E$?

I know that it's well defined and at least immersed:

Let $\pi: E \longrightarrow M$ a smooth fiber bundle and $f:M \longrightarrow N$ a smooth function. The Poor's book constructs the pullback bundle as the subset $f^*E:=\{(p,x) \in N\times E|\; f(p)=\pi(x)\}$ with projection $(f^*\pi)(p,x)=p$ and "trivialization charts":

$\Psi:(f^*\pi)^{-1}[f^{-1}[U]] \longrightarrow f^{-1}[U] \times F$


With $\psi:\pi^{-1}[U] \longrightarrow U\times F$, a bundle chart of $\pi:E \longrightarrow M$

I've already proofed that there exist a (unique) smooth structure in $f^*E$ such that $f^*\pi:f^*E\longrightarrow N$ is a smooth fiber bundle with trivialization charts $(f^{-1}[U],\Psi)$. By calculus in these charts, the inclusion $i:f^*E \longrightarrow N\times E$ is an immersion, so $f^*E$ is an immersed submanifold of $N\times E$.

Is it an embedded submanifold? Why? Is there an easy way to show that the inclusion is an embedding?


I found a post with the same question, the answer is interesting because it involves less structure, the minmum to apply transversality, but I think it has to exist a more easy way in the case. The Pullback Bundle is an Embedded Submanifold of its Parent Space

  • $\begingroup$ No, you have things backwards. If $E$ is a bundle on $M$, then you need a smooth function $f\colon N\to M$ (not the other way around), and then $f^*E$ will be the bundle on $N$ whose fiber over $p$ is the fiber of $E$ over $f(p)$. $\endgroup$ Dec 10, 2019 at 22:39

1 Answer 1


There is no secret. Let us start with definitions and some well-known facts.

Let $\phi : M_1 \to M_2$ be a smooth map between smooth manifolds $M_1, M_2$. It is called a (smooth) embedding iff $\phi(M_1)$ is a smooth submanifold of $M_2$ and $\phi : M_1 \to \phi(M_1)$ is a diffeomorphism. It is called an immersion if all derivatives $T_p \phi : T_p M_1 \to T_{\phi(p)} M_2$ are injective. Each immersions is locally an embedding (which means that each $p \in M_1$ has an open neighborhood $U$ such that $\phi\mid_U$ is an embedding). Immersions are in general no embeddings (even if they are injective). However, if an immersion is a topological embedding (which means that $\phi : M_1 \to \phi(M_1)$ is a homeomorphism), then $\phi$ is an embedding. See for example https://www.math.lsu.edu/~lawson/Chapter6.pdf.

Starting with the subspace $M_1 = f^*E$ of $M_2 = N \times E$, you have shown that the space $M_1$ can be endowed with a smooth structure which is uniquely determined by suitable requirements. You have moreover shown that the inclusion $i : M_1 \to M_2$ is an immersion. But $i$ is a topogical embedding by definition. Thus $f^*E = i(f^*E)$ is a smooth submanifold of $N \times E$.

  • $\begingroup$ Thanks, now I understand. For me $M_1$ was a subset, I defined its topology as the unique such that $\Psi$ were bundle charts. I see that the functions $\Psi$ and $\Psi^{-1}$ are continous putting the subspace topology in $M_1$. So the previuos topology I've found is the subspace topology and the inclusion is a topological (and smooth) embedding. $\endgroup$ Dec 15, 2019 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.